Game of Primes （博弈）

思路：

• 根据博弈条件: 能够转化到必败的条件,必赢,不然必输
• 然后另外一个技巧 看能不能后手一直让某一个条件维持不变,从而让自己获胜

1 初始情况可能要特判，更具自己的代码

Alice and Bob always like playing games with each other and today they found a new game about primes.

There are two positive integers xx and yy in the game, and Alice and Bob move in turn. At each turn, the current player can choose one integer and subtract it by 11 (making (x, y)(x,y) to (x - 1, y)(x−1,y) or to ( x, y - 1)(x,y−1)). The game ends when one of following conditions is met and the winner is specified at the same time:

When xx or yy equals to KK: Bob wins.
When xx and yy are both primes: Alice wins.
When both of the previous conditions are satisfied at the same time: Bob wins.
Now xx, yy, KK and who moves first are given, can you determine who will finally win the game if they both play optimally?

Input
The first line of input contains an integer TT, representing the number of test cases. Then following TT lines and each line contains one test case.

For each test case, there are four integers xx, yy, KK and ww separated by exactly one space. xx,yy,KK are mentioned above. w=0w=0 when Alice moves first and w = 1w=1 when Bob moves first.

Output
For each test case, you should output Case xx: name in one line, where xx indicates the case number starting from 1, and name is the player who will win the game.

Sample 1
Inputcopy    Outputcopy
4
4 9 2 0
7 10 2 0
6 39 2 0
5 28 2 0
Case 1: Alice
Case 2: Alice
Case 3: Alice
Case 4: Bob
Note
1 \le T \le 1001≤T≤100

2 \le x, y \le 10^62≤x,y≤10 
6
 

2 \le K \le \min(x, y)2≤K≤min(x,y)

0 \le w \le 10≤w≤1

For 90\%90% test cases: \max(x, y) \le 1000max(x,y)≤1000

#include <bits/stdc++.h>
using namespace std;
#define ri register int
#define M 1000005

template <class G> void read(G &x)
{
    x=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    x=f?-x:x;
    return ;
 } 
 

int flag[M],q[M];
void init()
{
    int r=0;
    for(ri i=2;i<=1e6;i++)
    {
        if(!flag[i])
        {
            q[++r]=i;
        }
        for(ri j=1;j<=r;j++)
        {
            if(i*q[j]>1e6) break;//
            flag[i*q[j]]=1;
            if(i%q[j]==0) break;
        }
    }
}
int k;
bool ck(int x,int y)
{
    while(x>k&&y>k)
    {
        if(!flag[x]&&!flag[y])
        {
            return 1;
        }
        x--;y--;
    }
    return 0;
}
bool pd(int x,int y)
{
            if(ck(x,y))
            {
                return 1;
               
            }
            if(ck(x-2,y)&&ck(x,y-2)) // attention
            {
                return 1;
               
            }
            return 0;
            
}
int T,x,y,w;
int main(){
    
    read(T);
    init();
    int tot=0;
    while(T--)
    {
        tot++;
        read(x);read(y);read(k);read(w);
        if((x==k||y==k&&!flag[x]&&!flag[y]))
        {
            printf("Case %d: Bob\n",tot);
            continue;
        }
        if(!flag[x]&&!flag[y])
        {
            printf("Case %d: Alice\n",tot);
            continue;
        }
        if(x==k||y==k)
        {
            printf("Case %d: Bob\n",tot);
            continue;
        }
        if(w==1)
        {
            if(pd(x,y))
            {
                printf("Case %d: Alice\n",tot);
            }
            else printf("Case %d: Bob\n",tot);
            continue;
        }
        if(w==0)
        {
            if(pd(x-1,y)||(pd(x,y-1)))
            {
                printf("Case %d: Alice\n",tot);
            }
            else printf("Case %d: Bob\n",tot);
            continue ;
        }
    }
    return 0;
    
    
    
}